The search for absolute infinity

[phoenixthoth]

i edited my post but it's not of real consequence now.

Originally posted by matt grime
Erm, in what way did you not understand the counter example to the 'proof' you've got? (the unversal thing isn't at issue here, just the assertion that as {a} is in P(X), that a must be
*an* element of X. a is a collection of elements of X is all that you can deduce. X is an element of P(X) yet X is not in general an element of X!

ok. let z be any set and consider P(x). suppose z is in P(x). is z a subset of x or not? suppose it is. then that means all elements of z are in x. ok? now take z={a}. since the assumption is that P(x) is the universal set, {a} is in P(x). then all elements of {a} are in x. that means a is in x.

Second. You what? By construction the map is injective, find distinct x and y with f(x)=f(y) for f the function defined in my last post. It's elementary to show that it is injective, unless you aer going to argue that I cannot split the universal set into those elements in the inductive set and those not.

let me get this straight. is this an equivalent example:
U=N union (U\N).
let f be a self mapping of U such that
f(x)=x+1 for x in N and
f(x)=x for x in U\N.
is that your example?

thanks for helping me correct my paper, btw.

 

[matt grime]

Originally posted by phoenixthoth
i edited my post but it's not of real consequence now.


ok. let z be any set and consider P(x). suppose z is in P(x). is z a subset of x or not? suppose it is. then that means all elements of z are in x. ok? now take z={a}. since the assumption is that P(x) is the universal set, {a} is in P(x). then all elements of {a} are in x. that means a is in x.

No, the assumption about the unversality of P(X) does not come into it. Firstly, {a} is a set with one element, a, that a is also a set is misleading. You cannot conclude that all the elements of {a} are in x. Secondly, the last line exactly states the objection that a is only a subset of x. You are confusing 'is a subset of a set' with 'is an element of a set of sets'

let me get this straight. is this an equivalent example:
U=N union (U\N).
let f be a self mapping of U such that
f(x)=x+1 for x in N and
f(x)=x for x in U\N.
is that your example?

thanks for helping me correct my paper, btw.

Not quite, I want a collection of sets labelled by N, not the set N itself. The axiom of infinity means that such must exist in any model we're looking at. The sets are:

0 - {} the empty set
1 -{{}} the set containing the empty set
2 -{{},{{}}} the set containing the two previoius sets.


I can just shift the labels by one here and leave all other sets unchanged.

 

[matt grime]

Acutally ignore the universal set power set unique thing for now, maybe some light has just come on in my head.

As written your proof could do with explanation, well, as written the 'proof' is wrong or at least the assertion needs more explaining, but the result might hold.

I believe you want to consider the set which contains the set which contains a, for then {a} is a subset of X, but it contains one element, then a is in X - too few braces used

 

[phoenixthoth]

"Not quite, I want a collection of sets labelled by N, not the set N itself. The axiom of infinity means that such must exist in any model we're looking at. The sets are:

0 - {} the empty set
1 -{{}} the set containing the empty set
2 -{{},{{}}} the set containing the two previoius sets.


I can just shift the labels by one here and leave all other sets unchanged."

i'm only detecting the essence of what you're saying but I'm not getting it just quite yet. the claim, your claim, is that there is a bijection between U and a proper subset of U. are we discussing this corollary: if x is a proper subset of U, then there is no 1-1 map from U to x? I've lost track because I'm having a brain fart and you shot off two points before i could handle the first one. overload! so, if we're discussing that corollary, then your example should indicate (hopefully as explicitly as i need it to be if possible) a set x that isn't U such that there is a 1-1 map from U to x. i think your claim is that there is a 1-1 map from U to U\{{}}. i do want to keep the axiom of infinity (especially since i think it might be a consequence of the universal set axiom and so i must keep it), so i want to exactly pinpoint my error. well, i admire you if this example is 'trivial'.

 

[matt grime]

First, the P(X) thing can be corrected, as hopefully you saw above:given any set Z, {{Z}} is in P(X)

so the set with one element {Z} is a subset of X, so Z is an element of X.



The second.

Yes, it is your corollary that there is no 1-1 map form U to a proper subset of U, and the statement that for all sets A, U\{A} is not in bijection with U. Well, that isn't true if you have the axiom of infinity.

 

[phoenixthoth]

doesn't {{Z}} is in P(X) imply that {{Z}} is a subset of x?

 

[matt grime]

Originally posted by phoenixthoth
doesn't {{Z}} is in P(X) imply that {{Z}} is a subset of x?

Yes. It implies that {{Z}} is a set of some things in X, right? ie that {Z} is some set of elements of X, but {Z} has only one element, Z, so Z must be an element of X. Careful with your bracketing.

 

[phoenixthoth]

"Yes."

so if {{Z}} is a subset of x then the following conditional holds for all sets y:
if y in {{Z}} then y in x. ok?

suppose y in {{Z}}.

first of all, that means y in x.

second of all, that means that y={Z}. this works the other way: {Z} is in {{Z}}.

hence, {Z} in x.

now go back to {a} is a subset of x (which follows from {a} is in P(x)):
for all sets y, if y in {a} then y in x. ok?

suppose y is in {a}.
1. y=a
2. y in x.
3. therefore, a in x.

it's ok as it is.

 

[matt grime]

Maybe it's me that's got his braces wrong, it's a headache, but i agree the result is true. I don't dispute the result, and I increasingly think I agree with your original proof.

I stand by the second issue though, about proper subsets

 

[phoenixthoth]

i will heed your advice about being careful though.

i'd still like you to be more specific and concrete with me on your counterexample because that would seriously damage the paper. i mean to be as detailed as possible because I'm not an expert in set theory so i can't understand your sketch.

 

[matt grime]

By the ZF axioms there is the empty set {}, then there is the set containing it { {} }, and then the set containing those { {} , {{}} }, and so on, each of these can be labelled by an element of N correspeonding to the cardinality. The axiom of infinity states, that when I say 'and so on' that actually there is an infinite number of sets created inductively (this apparently does not follow from all the other axioms) with the labels from all the natural numbers. If you have this, your universal set cannot be finite, and must contain these sets.


Since there is a bijection from N to a proper subset of itself (n to n+1) then there is a bijection from those sets to a proper subset of the sets, and defining it to be the identity for all other sets gives a contradiction to your corollary.

 

[phoenixthoth]

thank you.

now another question.
that corollary was meant to be the contrapositive to the statement: if f is a 1-1 function from U to x, then U = x. what's the contrapositive of that? because if the contrapositive is wrong that means the theorem it draws its energy from is wrong.

 

[phoenixthoth]

what are "those sets " you mention in paragraph 2? they're indexed by N right? but what are they? are they N, P(N), P(P(N)), etc., which can be indexed by 0, 1, 2, ...? just for reference, here is the axiom of infinity:
\exists x\left( \emptyset \in x\wedge \forall y\in x\left( y\cup \left\{ y\right\} \in x\right) \right)

 

[matt grime]

I said above

0 ---{} the empty set
1---{ {} } the set containing the empty set
2 ---{ {} , { {} }} the set containg the previous two sets

3 is the set continaing the previous 3 sets, and so on n is the set containing all the previous n-1 sets constructed. This is often called omega, a quick google for axiom of infinity wil provide you with some good links.

 

[phoenixthoth]

omega is the ordinal number for the set you're describing, which is N.

0 is defined to equal Ø.
1 is defined to equal {Ø}.
2 is defined to equal {0,1}
3 is defined to equal {0,1,2}.
N is defined to equal {0,1,2,...} which exists by the infinity axiom.
do a search on that yourself. see enderton's "elements of set theory," et al.

so your map really is this:
f(x)=x+1 for x in N
f(x)=x for x in U\N.
that's what i thought.

 

[phoenixthoth]

a point of clarification: when we write U\N or any relative complement, do we mean the set of elements in U that are not in N or the set of elements such that it is not true that they are in N?

ie, U\backslash N=\left\{ x\in U:x\notin N\right\} or
U\backslash N=\left\{ x\in U:\lnot \left( x\in N\right) \right\} =\left\{ x\in U:x\in _{M}N\vee x\notin N\right\}?

 

[phoenixthoth]

Originally posted by phoenixthoth
a point of clarification: when we write U\N or any relative complement, do we mean the set of elements in U that are not in N or the set of elements such that it is not true that they are in N?

ie, U\backslash N=\left\{ x\in U:x\notin N\right\} or
U\backslash N=\left\{ x\in U:\lnot \left( x\in N\right) \right\} =\left\{ x\in U:x\in _{M}N\vee x\notin N\right\}?

which of these is true, if any:
\left\{ x\in U:x\in _{M}N\vee x\notin N\right\} \cup N=U and/or
\left\{ x\in U:x\notin N\right\} \cup N=U

 

[matt grime]

Originally posted by phoenixthoth
omega is the ordinal number for the set you're describing, which is N.

0 is defined to equal Ø.
1 is defined to equal {Ø}.
2 is defined to equal {0,1}
3 is defined to equal {0,1,2}.
N is defined to equal {0,1,2,...} which exists by the infinity axiom.
do a search on that yourself. see enderton's "elements of set theory," et al.

so your map really is this:
f(x)=x+1 for x in N
f(x)=x for x in U\N.
that's what i thought.

Some people label the set omega, some N, whatever. It is just a label. Personally, I would never say a number is *equal* to a set, but then I don't define my numbers as sets, cos I don't do set theory.

It's your set theory, I don't know what your definition of complement is. As said originally, perhaps you've axiomatized these issues away. Perhaps there are things which you do not know to be subsets, this is your three valued logic system, you ought to know what is what. My comments are pointing out where there are possible issues. I owuld suggest that I'm defining f(x)=x for all sets X where the value of X in N is NOT T, which I naively assume to be F OR M. Presumably it is not possible for a statement to be simultaneously T AND M, that is T AND M is F.

 

[phoenixthoth]

"It's your set theory, I don't know what your definition of complement is. As said originally, perhaps you've axiomatized these issues away. Perhaps there are things which you do not know to be subsets, this is your three valued logic system, you ought to know what is what. My comments are pointing out where there are possible issues. I owuld suggest that I'm defining f(x)=x for all sets X where the value of X in N is NOT T, which I naively assume to be F OR M. Presumably it is not possible for a statement to be simultaneously T AND M, that is T AND M is F."

i guess what i was asking was what do you think is the best way to define complements to cover all the loose ends. you're right i should know that and i'll look into it. i am not sure that certain things we're talking about are sets anyway so it may not matter, or they may be fuzzy sets so it may be important to recognize that. i believe N is a crisp set and so how one defines complement doesn't matter. that's my gut feeling. hmm... your naivete is in your mind for i did say that veracity thingy's are functions which entails that they can't have simultaneously values T and M which is to say that T AND M is F. a veracity relation would be a whole different story and wouldn't resemble normal set theory at all in any way shape or form.

brain fart and for objective verification: what is the contrapositive of the statement "if f is a 1-1 function from U to x, then x=U?"

second: that is theorem 3. do you care to point out the error in theorem 3? i just want to pin down exactly where it's incorrect and see if it's repairable. now the whole theory wouldn't crumble if theorem 3 had to be torn out but i found it rather nice to have it there because it meant nothing i could think of was bigger than U. hmm... well, one doesn't need theorem 3 for that i see now. you know what? i had spotted a major error in my paper a while back and wondered if anyone else would notice it but this wasn't it. however, the major error was a consequence of theorem 3. good thing 3 isn't essential. but it will mean another push is ahead of me. I'm actually glad to see this now because i really hated what this major error had to say anyway. the major error implied that ALL sets are FUNCTIONS which is FALSE! that was the major error i wanted to see if others noticed. it is, in fact, corollary 4 to theorem 3. you almost got to it as you noticed a problem with theorem 3 in corollary 2. sigh... ok. it can still work though i will have to change some things around.

thanks again for your patient feedback.

 

[matt grime]

Well, my issue so far is with the part of the proof the theorem three that states

g: D to R ... if g^-1(y)...

well, g is only injective, therefore one can only define the inverse for the y in the image, that is y cannot be arbitrary, and here D and R are any sets and G any injection. So you may not define the preimage of elements not in the image.

I didn't read as far as the corollary you found to be in error.

 

[phoenixthoth]

i always had issues with that proof. i just didn't believe it 100%.

well, so much for theorem 3!

now i got to try to refit the carpet into the room...

 

[matt grime]

Just use a pair of Banach-Tarski scissors, it'll always fit

 

[phoenixthoth]

LOL

if this ever gets into a publishable form, i will definitely thank you and hurkyl for the feedback in it. I've been wondering who i'd dedicate it to. i think i may dedicate it to the letter M.

 

[matt grime]

Here's a thought. Seeing as you are axiomatically assuming a universal set anyway, part of the definition of its universality ought to be that if f is any injection from U to S, some set S, then f must be a bijection. I think the problem here is that you are attempting to prove what an axiom without using the relevant axioms. There is no harm in assuming this as an explicit axiom, and it might be that with a little thinking there is some way round this. Here I would suggest that the issue is resovable, by saying IF f is an injection, then S must be 'at least as big' as U, as U is universal then if f is not surjective, you have some problem (phrase it in your own preferred manner of thinking). Because really, the issue is that if f is a bijection from U to S that S is equal to U. Note that the set of natural numbers is bijective with the rationals but they are not EQUAL. Perhaps now the problems vanish a little.

 

[phoenixthoth]

are you trying to get on the co-authors list now? you're more than welcome to! so you want a tenneson number of 1, huh? maybe one day it will be time to stamp "three truth values are sufficient" on our stationary LOL.

hmm... i'd hate to add another axiom but i will if i must.

are you saying that U in bijection with x SHOULD imply U=x or SHOULD NOT imply U=x? i no longer think it should. i do think that if f is an injection from U to x then there exists a bijection between U and x.

 

[matt grime]

Not hankering for a co-authorship in the slightest.

It seems that there is some lattitude in what one means be universal here, and the definition of complement may allow for it.

Clearly if F:U to S is injective then, since the natrual inclusion S to U is injective, it is a 'proof from the book' that U and S are bijective, at least in ordinary binary logic.

 

[phoenixthoth]

of course! cantor-schroeder!

i was just trying to prove that theorem and i was like banging my head against the wall. almost got it (NOT) but i decided to see if you had the easier way and you did. my teacher always told me to not get the office next to the library because you have to think about it rather than look it up and i cheated and cheat by asking you. how real mathematicians go it alone is unimaginable ;).

no seriously, you can be a co-author if you want. in fact if you have more letters after your name than myself then that could add credibility to it. not saying it's publishable now or that anyone would care to read it but one day it will be doable. one day soon (like three months max i suspect, depending on when i get around to re-vamping it which I'm now a lot more eager to do than i was half an hour ago)...

 

[phoenixthoth]

version 9

here it is:
http://www.alephnulldimension.net/matharticles/tuzfcver9.pdf

any thoughts?

 

[matt grime]

What does the wavy equals sign mean here? (Yes, theorem 3!)

 

[phoenixthoth]

i should probably specify it in the document in the next draft...

in bijection with.